Nuclear and Particle Physics - Lecture 11 Parity and charge conjugation conservation 1 Introduction We have seen the interaction vertices for QED and QCD and know that these do not change the types of particles, i.e. the numbers of quarks and leptons. As we will see, the weak interaction also conserves all these quantities. QED and QCD also conserve the fermion flavour, meaning for quarks u, d, etc, and for leptons e, µ, etc. It turns out the weak force does not conserve flavour. Another important property of the weak interaction is that it does not conserve the parity (P ) or charge conjugation (C) quantum numbers. Let's look at what P and C conservation gives us in strong and EM interactions and how we know it is not conserved in some weak processes. 2 Symmetries and conservation laws There is a very deep connection in physics between symmetries and conserved quantities. For example, we can do an experiment today or tomorrow. Assuming we remove all the external factors which depend on when we do it, we should get the same answer. This is because there is a time translation symmetry; there is no absolute time scale. This symmetry actually leads to conservation of energy. Similarly, symmetry under translation in each of three directions in space leads to conservation of the three components of momentum. Symmetry of direction in space (as space is isotropic) leads to conservation of angular momentum. Parity and charge conjugation are further examples. Parity is an operation which takes a vector, e.g. the space position r and reflects it through the origin to make it −r. Charge conjugation takes every particle and replaces it with its antiparticle (and vice versa). It seems obvious from everyday experience that things will work the same after a parity operation; a mirror-image of any machine will function in the same way. It is maybe not so obvious that we would expect systems to be the same after swapping particles and antiparticles; however, if you think about the EM force between any pair of particles, it goes as Q1Q2 so swapping the signs of Q1 and Q2 in fact keeps all the forces the same and since the masses are the same, this has no effect. Parity and charge conjugation are somewhat different from the other symmetry cases because they are not continuous symmetries, but are discrete. Take the example of time translation invariance which leads to energy conservation. Because any translation is possible, then any energy is allowed. It turns out that for a discrete transformation symmetry, then there is a discrete quantum number which is conserved. Hence, there is still a connection between a symmetry and a conserved quantity, even for the discrete case. What are these discrete eigenvalues for parity and charge conjugation? Consider the case for the parity operator, P^, which reflects polar vectors through the origin r ! −r. For an eigenstate of parity, where P^ = P it is obvious that applying the parity operation a second time must return everything to its original state, so P^2 = P P^ = P 2 and since this must be just again, then P 2 = 1 and the eigenvalues of parity must be 1. The same argument holds for the charge conjugation operator; this changes particles to antiparticles 1 so a second application must return everything back to the original state. Again, the eigenvalues of C^ must be C = 1. Hence, we can look at parity (or charge conjugation) conservation either in terms of the whether there is a symmetry or in terms of whether the quantum number of P or C is conserved. It is a little unfortunate that the same word, \parity", is used to mean both the symmetry and the conserved quantity. Compare to time translation (the symmetry) and energy (the conserved quantity), which never get confused. It will hopefully usually be clear from context which is meant. It is often hard to see a situation in terms of both views, so we will use either the symmetry or the conserved quantity, depending on which is easier, for any case. What we find is that QED and QCD conserve P and C, or equivalently respect P^ and C^ symmetries, whereas the weak force does not. We will first consider some cases where we can understand things easily in terms of the conserved quantum numbers and later in terms of symmetries. 3 Conserved quantities; decays of mesons As was stated in a previous lecture, of the nine ground state meson flavour combinations, only the three flavour neutral states, i.e. those made out of combinations of uu, dd and ss, specifically the π0, η and η0, can decay electromagnetically. This interaction does not change flavour and so the others, such as ud, have no lower mass state to go to. The excited states, in contrast, can all decay strongly as there is always at least one lower mass state of the same net flavour. The mesons all have well-defined parity values which turn out to be −1, i.e. they are parity eigenstates P^jπ+i = −|π+i Along with being spin 0, this is why they are written as J P = 0−. In contrast, only the flavour neutral mesons are C^ eigenstates as these are the only ones which are their own antiparticles. For the others, for example the charged pion C^jπ+i = C^judi = judi and so is clearly not a C^ eigenstate. However, the three flavour neutral ground state mesons are their own antiparticles and for them, it turns out that C = +1, i.e. C^jπ0i = +jπ0i Hence both P and C are well defined so they are written as J P C = 0−+. This information comes from P and C conservation in the flavour neutral meson decays so let's take a look at how this works. Consider the π0 first. The π0 decays almost 99% of the time to two photons; this is obviously an EM decay. Comparing the Feynman diagrams for two and three photon decays q e q e q γ q γ eq γ γ γ eq eq q q 2 then we might think the three photon decay has an extra vertex and hence the decay rate will be down by a factor of α ∼ 1=137, which accounts for the other 1% of the decays. This is not the case. The photon is its own antiparticle so it is also a C^ eigenstate. Its charge conjugation value is Cγ = −1, which is clear if you think of an arbitrary system of charges with their resulting E and B fields. Under a charge conjugation operation, all the charges become the opposite and so all the field values becomes reversed E ! −E; B ! −B The photon consists of these fields, so it must be that C^jγi = −|γi This means a system of n photons has an overall C value of (−1)n. Hence, two (or any even number of) photons have Cγγ = +1 while three (or any odd number of) photons have Cγγγ = −1. Thus, the three photon π0 decay would change C from +1 to −1, which is not allowed for an EM decay as it is a conserved quantity. In fact, the three photon decay is not 1% of the two photon decay, but has never been observed and is known to be less than 10−7 of the two photon rate. (As an aside, going back to positronium in lecture 5, C conservation is also the reason why each of the hyperfine split ground states only goes to two or three photons.) Now consider the η. For identical reasons to the above, it can decay to two photons but not + − three. However, since mη > 3mπ, it is also heavy enough to decay to two pions, e.g. π π , or three pions, e.g. π+π−π0. However, the former does not occur. Consider the two pion decay; the parity for any orbital angular momentum state is the same as its exchange symmetry, i.e. L P − PL = (−1) , and the intrinsic parity of the pions is Pπ = −1 (i.e. J = 0 ), so the total parity L of a two pion state is P = PπPπPL = (−1) . The η and the pions have no spin, so L must be zero for this decay. Hence for this decay, the two pion final state parity would be PπPπPL = +1, while the η itself has intrinsic parity, like the pions, of Pη = −1. Obviously, adding a third pion but keeping L = 0 gives a total P of PπPπPπPL = −1 which is the same as the η. Hence, the two pion decay would violate parity conservation, whereas the three pion decay would not. The two pion decay has never been observed; the experimental limit on the rate for two pion decays is at most 10−3 of the three pion rate. In fact, this argument is valid for any of the ground state, J P = 0−, meson decays as they all have well-defined parity; none of them can decay to two pions (or indeed any two of the ground state mesons) without violating parity. In fact, the first decay observed which violated parity was the decay K+ ! π+π0, which has a rate roughly three times that of the three pion decay. This cannot be a strong or EM decay as the strange quark disappears; it is indeed a weak decay. At the time, the significance of this was not understood and it was thought there were two kaon-like particles of the same mass but with different intrinsic parities, one of which decayed to two pions and one to three pions.